AMC12 2022 B

AMC12 2022 B · Q9

AMC12 2022 B · Q9. It mainly tests Sequences & recursion (algebra), Primes & prime factorization.

The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that \[2^{a_7}=2^{27} \cdot a_7.\] What is the minimum possible value of $a_2$?

数列$a_0,a_1,a_2,\cdots$是一个严格递增的正整数等差数列，满足 \[2^{a_7}=2^{27} \cdot a_7.\] $a_2$的最小可能值是多少？

(A) 8 8

(B) 12 12

(D) 17 17

(E) 22 22

Answer

Correct choice: (B)

正确答案：（B）

Solution

We can rewrite the given equation as $2^{a_7-27}=a_7$. Hence, $a_7$ must be a power of $2$ and larger than $27$. The first power of 2 that is larger than $27$, namely $32$, does satisfy the equation: $2^{32 - 27} = 2^5 = 32$. In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so their graphs will not intersect again. Now, let the common difference in the sequence be $d$. Hence, $a_0 = 32 - 7d$ and $a_2 = 32 - 5d$. To minimize $a_2$, we maxmimize $d$. Since the sequence contains only positive integers, $32 - 7d > 0$ and hence $d \leq 4$. When $d = 4$, $a_2 = \boxed{\textbf{(B)}\ 12}$.

我们可以将给定方程重写为$2^{a_7-27}=a_7$。因此，$a_7$必须是2的幂且大于27。第一个大于27的2的幂，即32，满足方程：$2^{32 - 27} = 2^5 = 32$。事实上，这是唯一解；$2^{a_7-27}$是指数的，而$a_7$是线性的，它们的图像不会再次相交。现在，设数列的公差为$d$。故$a_0 = 32 - 7d$，$a_2 = 32 - 5d$。要最小化$a_2$，我们最大化$d$。由于数列只包含正整数，$32 - 7d > 0$，故$d \leq 4$。当$d = 4$时，$a_2 = \boxed{\textbf{(B)}\ 12}$。

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